Given three positive numbers $a,\,b,\,c$ . Prove that $(\!abc+ a+ b+ c\!)^{3}\geqq 8\,abc(\!1+ a\!)(\!1+ b\!)(\!1+ c\!)$ .

My own problem is given a solution, and I'm looking forward to seeing a nicer one(s), thank you !

Solution. Without loss of generality, we may suppose $(a- 1)(b- 1)\geqq 0 \therefore 1+ ab\geqq a+ b$. Then

$$(abc+ a+ b+ c)^{3}= \left ( c(1+ ab)+ a+ b \right )^{2}\left ( c(1+ ab)+ a+ b \right )\geqq$$

$$\geqq 4\,c(1+ ab)(a+ b)\left ( c(a+ b)+ (a+ b) \right )= 2\,c(a+ b)^{2}(2+ 2\,ab)(1+ c)$$

Again, by using a.m.-g.m.-inequality, we have $2c(\!a+ b\!)^{2}(\!2+ 2ab\!)(\!1+ c\!)\geqq 8abc(\!1+ a\!)(\!1+ b\!)(\!1+ c\!)$



After replacing $a\rightarrow\frac{1}{a},$ $b\rightarrow\frac{1}{b}$ and $c\rightarrow\frac{1}{c}$ we need to prove that $$(1+ab+ac+bc)^3\geq8abc(1+a)(1+b)(1+c).$$ Now, let $a+b+c=3u$, $ab+ac+bc=3v^2,$ where $v>0$ and $abc=w^3$.

Thus, we need to prove that: $$(1+3v^2)^3\geq8w^3(1+w^3+3v^2+3u)$$ or $$(1+3v^2)\geq8(w^3+w^6+3v^3w^2+3uw^3)$$ and since by AM-GM $v\geq w$ and $v^4\geq uw^3,$ it's enough to prove that: $$(1+3v^2)^3\geq8(v^3+v^6+3v^5+3v^4)$$ or $$(1+3v^2)^3\geq8v^3(1+v)^3$$ or $$1+3v^2\geq2v(1+v)$$ or $$(1-v)^2\geq0$$ and we are done!

Also, $uvw$ kills this inequality, but it's not so nice.

  • $\begingroup$ YOU'RE SO SUPER ! What makes you think to replace by a substitution? $\endgroup$ – user680032 Jun 23 at 5:15
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    $\begingroup$ I think your solution is the neatest and the best one ! $\endgroup$ – user680032 Jun 23 at 5:15
  • $\begingroup$ @HanhVyheaven28 Because after this substitution the inequality looks simpler.Thank you! $\endgroup$ – Michael Rozenberg Jun 23 at 5:16
  • $\begingroup$ Wow, that's so impressive! I found it by discriminant and you do this easily! $\endgroup$ – user680032 Jun 23 at 5:18

Hint: By AM-GM, $$(abc+a) + (b+c) \geqslant 2a\sqrt{bc}+2\sqrt{bc} = 2\sqrt{bc}(1+a)$$

Now multiply three such inequalities (after cyclical shift).

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    $\begingroup$ Very nice! Bravo! +1 $\endgroup$ – Michael Rozenberg Jun 23 at 7:38
  • $\begingroup$ 'Now multiply three such inequalities (after cyclical shift).', that made a big impres ! $\endgroup$ – user680032 Jun 23 at 9:12

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